Could anyone give me a couple specific examples of how sections of a Presheaf on discrete topology would or could glue together? If I am correct, it depends on the mapping one defines. Now I have worked out an example in which I define a presheaf $\mathcal{F} = Fct(Op_X^{op},\mathbf{Set})$ as follows:
Let $X = \{a,b,c,d\}$ be a set and $\mathcal{T}_X$ be the discrete topology on $X$. Say I defined the presheaf on the open sets of $\mathcal{T}_X$ as the $\mathcal{F}(X) = \{0,1,2,3\}$, the presheaf on all sets of 3 elements i.e. $\mathcal{F}(\{a,b,d\}) = \{0,1,2\}$. The presheaf on all sets of 2 elements i.e. $\mathcal{F}(\{a,d\}) = \{0,1\}$, and the presheaf on singletons such as $\mathcal{F}(\{b\}) = \{0\}$ as well as $\mathcal{F}(\{\emptyset\})=\{0\}$.
Now a presheaf is a functor $\mathcal{F}$ on $X$ which associates to each open set $U \subset X$, a set $\mathcal{F}(U)$, and to an open set $V \subset U$ a map $\rho_{UV}: \mathcal{F}(U) \rightarrow \mathcal{F}(V)$ such that for each open inclusion $W \subset V \subset U$ one has $\rho_{UV}\circ\rho_{VW}=\rho_{UW}$ and that $\rho_{UU} = id_U$.
So as $\{b\} \subset \{a,b\} \subset \{a,b,d\}$ (for only one of the possibilities), I would have to make sure that as $\rho_{UV}: \mathcal{F}(\{a,b,d\}) \rightarrow \mathcal{F}(\{a,b\})$ and $\rho_{VW}: \mathcal{F}(\{a,b\}) \rightarrow \mathcal{F}(\{b\})$ in the order $W \subset V \subset U$ as above, that $\rho_{UW}: \mathcal{F}(\{a,b,d\}) \rightarrow \mathcal{F}(\{b\})$? Hence we would have by the set associated to each open set in $\{a,b,d\}$ for this example, that $\rho_{UV}: \{0,1,2\} \rightarrow (\{0,1\})$ and $\rho_{VW}: (\{0,1\}) \rightarrow \mathcal{F}(\{0\})$ that $\rho_{UW}: \mathcal{F}(\{0,1,2\}) \rightarrow \mathcal{F}(\{0\})$. Now it seems that by the way I defined the presheaves (and the sets associated with them), that this would be satisfied for all possible sets of inclusions.
The above was my attempt at defining a presheaf which I could turn in to a sheaf (sheafify?). $\textbf{Question}$ How do I glue these sections together so as to create a sheaf?
Here is my attempt at working out the "mechanics" of this, I am extremely interested in actually physically seeing how these sections fit together - more than just texts saying there is a "compatible gluing", how is it compatible and why?
Say I took an open subset $U = \{a,b,d\} \in \mathcal{T}_X$. Now I find an open covering of $U = \bigcup U_{i}$ Let $U_{i} \in \{\{a,b\},\{d\}\}| i \in I$ where $I$ is an indexing set. So I have that $U = \{a,b\} \cup \{d\}$. Now if I understand correctly (and I am unsure of how to proceed), but say I take another open covering of $U$, say $U = \bigcup U_{j}$ where $U_{j} \in \{\{a\},\{b,d\}\}|j \in J$ where $J$ is the indexing set. Hence we could have that $U = \{a\} \cup \{b,d\}$. So I need that the sections which are restricted to all combinations of the intersections of individual subsets of both open coverings, agree. That being said if a section $s_i \in \mathcal{F}(U_i)$ equals another section $s_j \in \mathcal{F}(U_j)$ then $\exists s \in \mathcal{F}(U)$ such that $s|_{U_i}=s_i \, \forall i \in I$. So it seems as if we are choosing, for example $U_{i} \in \{\{a,b\},\{d\}\}$ and $U_{j} \in \{\{a\},\{b,d\}\}|j \in J$ We are wanting to see if there is an agreement between sections on this intersection i.e. $s_{i}|_{U_{ij}} = s_{j}|_{U_{ij}} \, \forall i,j$. So with the example sets as $U_{ij} = U_i \cap U_j$
First, denoting the two possible set of sections above as $s_i$ as $s_{a_i}$ and $s_j$ as $s_{b_j}$ so as to not mix up the various sections on the two different open coverings $U_i$ and $U_j$ of $U$ and letting $\{U_1 = \{a,b\}, U_2 = \{d\}\} = U_i$ and $\{U_1 = \{a\},U_2 = \{b,d\}\} = U_j$ For $U_{ij} = U_i \cap U_j$ we would get the set of intersections $\{U_{11},U_{12},U_{21},U_{22}\} = \{\{a\},\{b\},\emptyset,\{d\}\}$ respectively. Now if sections must agree then, again please let me know if I am on the right track but for presheaves to be glueable we have to have agreement on all sections and so $s_{a_i}|_{U_{ij}} = s_{b_j}|_{U_{ij}} \forall i,j$ we then have to have
$s_{a_1}|_{U_{11}} = s_{b_1}|_{U_{11}} \rightarrow s_{a_1}|_{\{a\}} = s_{b_1}|_{\{a\}}$
$s_{a_1}|_{U_{12}} = s_{b_2}|_{U_{12}} \rightarrow s_{a_1}|_{\{b\}} = s_{b_2}|_{\{b\}} $
$s_{a_2}|_{U_{21}} = s_{b_1}|_{U_{21}} \rightarrow s_{a_2}|_{\emptyset} = s_{b_1}|_{\emptyset}$
$s_{a_2}|_{U_{22}} = s_{b_2}|_{U_{22}} \rightarrow s_{a_2}|_{\{d\}} = s_{b_2}|_{\{d\}}$
Where of course due to how I $\textbf{defined}$ the presheaf maps I can look at agreement. So to have a section restricted to a particular element means?
for $s_{a_i} \in \mathcal{F}(U_i)$ we would have $s_{a_1} \in \mathcal{F}(\{a,b\}),s_{a_2} \in \mathcal{F}(\{d\})$ and $s_{b_1} \in \mathcal{F}(\{a\}),s_{a_2} \in \mathcal{F}(\{b,d\})$
In this context (if it is correct) what is a section? It seems as if it is some sort of morphism of morphisms meaning that somehow it is not only a morphism of opposite inclusions in the open sets of the category of open sets, but also a morphism which takes this morphism (of opposite inclusions) to a morphism of some structure of sets (for example) inside the Category of sets. If you are familiar with 2-D arrays in computer science this seems like a similar concept - an array of arrays. I don't know if it works describing it this way, but I am trying to understand the concept.
So what do these sections look like? Also perhaps someone could give a concrete example of a morphism via the way I defined it, that is not a compatible section?
That is a long question. I'm not very comfortable with all your notations ($U_i=\{U_1,U_2\}$ ? that is weird), and I don't understand why you consider two open coverings at the same time.
So let $\mathcal{F}$ be your presheaf you want to sheafify (assuming you've checked it is really a presheaf, which by the way may be very difficult if you define all the restriction maps "by hand"). Let's call $\mathcal{F}^a$ the associated sheaf and $U$ an open set.
So what is a section $s\in\mathcal{F}^a(U)$ ? If you have an open covering $(U_i)_{i\in I}$, then you know that the data of $s$ is equivalent to the data of $s_i\in\mathcal{F}^a(U_i)$ which agree on the $U_i\cap U_j$. But you need to know what the $\mathcal{F}^a(U_i)$, the $\mathcal{F}^a(U_i\cap U_j)$ and the restriction maps are.
Now, if the topology is discrete, there is an obvious covering that make the whole thing easily computable : an open set $U$ is the union of the singletons $\{x\}$ for $x\in U$. Because the intersections are empty, the sheaf condition reads : the data of $s\in\mathcal{F}^a(U)$ is equivalent to the data of $s_x\in\mathcal{F}^a(\{x\})$ for $x\in U$.
Moreover, you can show that $\mathcal{F}^a(\{x\})=\mathcal{F}^a_x=\mathcal{F}_x=\mathcal{F}(\{x\})$ (where $\mathcal{F}_x$ is the stalk of $\mathcal{F}$ at $x$).
So we have proven : $\mathcal{F}^a(U)=\prod_{x\in U}\mathcal{F}(\{x\})$. In your case, $\mathcal{F}^a(U)$ will be a singleton for every open set $U$.